Dry air was passed successively through a sol | vedclass

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(A) According to Ostwald-Walker method,the relative lowering of vapour pressure is given by: $\frac{p^o - p}{p^o} = \frac{w 	imes M}{W 	imes m}$.Her

Vedclass · Accepted Answer

$70.31$

Vedclass · Answer

(A) According to Ostwald-Walker method,the relative lowering of vapour pressure is given by: $\frac{p^o - p}{p^o} = \frac{w 	imes M}{W 	imes m}$.Here,loss in weight of solution is proportional to $p$ and loss in weight of solvent is proportional to $(p^o - p)$.Given: weight of solute $(w) = 5 \ g$,weight of solvent $(W) = 80 \ g$,loss in solution $= 2.50 \ g$,loss in solvent $= 0.04 \ g$.Formula: $\frac{p^o - p}{p} = \frac{	ext{loss in solvent}}{	ext{loss in solution}} = \frac{0.04}{2.50}$.Also,$\frac{p^o - p}{p} = \frac{n}{N} = \frac{w/m}{W/M} = \frac{w 	imes M}{W 	imes m}$.Substituting values: $\frac{0.04}{2.50} = \frac{5 	imes 18}{80 	imes m}$.$m = \frac{5 	imes 18 	imes 2.50}{80 	imes 0.04} = \frac{225}{3.2} = 70.31 \ g/mol$.

Dry air was passed successively through a solution of $5 \ g$ of a solute in $80 \ g$ of water and then through pure water. The loss in weight of solution was $2.50 \ g$ and that of pure solvent $0.04 \ g$. What is the molecular weight of the solute?

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